Profit and Loss

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Solving (I) and (II) => From (i) C = 200000 -.H
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**Sample questions for
Placement Papers – Solved Problems 1**

3 mangoes and 4 apples costs Rs 85. 5 apples and 6 peaches costs
Rs. 122. 6 mangoes and 2 peaches cost Rs.114. what is the combined price of 1
apple,1 peach and 1 mango?

(a)Rs 37 (b)Rs 39 (c) Rs 35 (d)Rs 36 e)None

Answer
: A

Explanation
:

3
mangoes and 4 apples costs Rs 85 => 3m+4a=85...(i)

5
apples and 6 peaches costs Rs. 122 =>
5a+6p=122....(ii)

6
mangoes and 2 peaches cost Rs.114.=> 6m+2p=114....(iii)

On
solving (i) & (iii),we get the
equation in terms of a and p => -8a+2p=-56 ..(iv)

m=15,a=10,p=12

Now
solving (ii) and (iv) , we get 58p=696
=> p=12

Substituting
p=12 in equation (iii) , we get m=15

Substituting
m=15 in equation (i), we get a=10

m=15,a=10
and p=12

The
combined price of 1 apple , 1 peach and 1 mango = m +a+ p= 15 +10 +12=Rs37

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**Sample questions for Placement Papers – Solved Problems 2**

A person buys a brand new Honda bicycle for Rs.800. He uses it for a year and
then sells it. He makes a profit of 10% in this transaction. Then, again, he
buys a new cycle but finds that the price has gone up by 20%. He uses the new
cycle for a year and then sells it for a profit of 30%. Find the average annual
profit of the person through these transactions.

a)160 b)184 c)168 d)172

Answer: B

Explanation: He
bought for 800 at first.

He Sold at a profit
of 10% => Profit on first cycle = 800 $ \times\frac{10}{100}$=80

He again bought a
new one at a price hike of 20% =-> New price= 800 $ \times\frac{120}{100}$ =960

He sold the second cycle
at 30% profit = > Profit on the second cycle = 960 $ \times\frac{30}{100}$ =288

Total profit in 2
transaction is=80+288=368;

Average annual
profit i = $\frac{368}{2}$=184rs

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**Sample questions for Placement Papers – Latest Pattern Solved Problems 3**

A man purchased cow and horse at rs 200000 and
sold a cow for 20% profit and horse at 20% a loss, if the overall gain is 4000 find
the selling price of cow and horse?

a)
Rs 132000,Rs 72000 b)Rs 140000,Rs64000

c)Rs 114000,Rs 90000 d) Rs 124000,Rs 80000

Answer A

Explanation:

Let the price of a cow is 'C' and the price of the horse is
'H'

The cost price of cow and horse => C + H =200000 ....(I)

The cost price of cow and horse => C + H =200000 ....(I)

He sold the cow for 20% profit and horse at 20% loss.
In overall he gained Rs 4000

Selling
price of cow and horse => 1.2 C + 0.8 H =204000....(II)

Solving (I) and (II) => From (i) C = 200000 -.H

Substituting this in (II) , we get H = 90000 and
C= 110000

Cost price of horse = 90000 and cost prie of cow
=110000

Selling price of Cow = 110000 $\times\frac{120}{100}=132000$

Selling price of Horse = 90000 $\times\frac{90}{100}$=72000

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**Sample questions for Placement Papers – Latest Pattern Solved Problems 4**

If Suraj's sum of the salaries of 2003, 2004, 2005 is $36400 and
the salary gets incremented 20% every year then find the salary drawn by Suraj
in the year 2005?

a)
14400 b) 15200 c) 16500 d)18000

Answer
: A

The
salary of Suraj is increased by 20% every year.

When
a salary is increased by 20%, its value after increase is salary $ \times\frac{120}{100}$

= salary $ \times\frac{6}{5}$

In
2004 , his salary = S $ \times\frac{6}{5}$

In 2005 , his salary = S $ \times\frac{6}{5} \times\frac{6}{5}$

Sum
of his salaries in 2003,2004 and 2005 =>$ S \;+\,\frac{6}{5}S\,+ \;\frac{36}{25}S\;$= $36400

On solving, we get
S=10000

Salary
in 2005 = 10000 $\times\frac{120}{100}\times\frac{120}{100}$
=
$14400

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**Sample questions for Placement Papers – Solved Problems 5**

A man sold 12 candies at 10 dollars had a loss of b% and again sold 12 candies at 12 dollars had a profit of b%. Find the value of b?

a)10% b)9.09% c)8% d)11.11%

Answer:
B

Explanation:

Here
the profit and loss in both transactions is the same when they are sold at two different
prices, their Cost price will be the average of two selling prices.

CP=$\frac{12+10}{2}$

Therefore the cost price is $11

Now

11 $ \times\frac{100+b}{100}$ =12

=9.09%

**Alternate Method :**

Let
the cost price be CP

One sold at b% profit and other sold at b%
loss

Loss $ =\frac{CP-SP}{CP}\times100\implies\;\frac{CP-10}{CP}\times100%
=b\% $
----------->1

Profit $=\frac{SP-CP}{CP}\times100\implies\;\frac{12-CP}{CP}\times100 %
=b\% $
------->2

Here both profit and loss are equal

$\frac{CP-10}{CP}\times100=\;\frac{12-CP}{CP}\times100 $

On solving , we get CP=11

Now
substituting the value of CP in equation => $\frac{11-10}{11}
\times100 $ %= b%

=
> b%= 9.09%

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**Sample questions for Placement Papers – Solved Problems 6**

A dealer buys a product at Rs.1920. he sells at a discount of
20% still he gets the profit of 20%. what is the selling price?

a)Rs. 2304 b) Rs. 1536 c) Rs. 2200 d)
it is not possible

Answer
: A

Explanation
:

Cost
price of the product is Rs 1920.

Discount
always given on marked price and profit is always obtained on the cost price

$ CP\, \times\frac{120}{100} =\, MP\, \times \frac{80}{100}$
(20% profit got
even after giving 20% discount on MP)

$1920\, \times\frac{120}{100} =\, MP\, \times \frac{80}{100} $

MP=2880

Selling
price = $MP\, \times \frac{80}{100}\, =\, 2880\times\frac{80}{100} $= Rs 2304

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**Sample questions for Placement Papers –Latest Pattern Solved Problems 7**

A Person buys a horse for 15 pounds, after one
year he sells it for 20 pounds. After one year, again he buys the same horse at
30 pounds and sells it for 40 pounds. What is the profit for that person?

option

(a) 20 pounds (b) 15 pounds (c) 25 pounds (d) 10 pounds

Answer : b

option

(a) 20 pounds (b) 15 pounds (c) 25 pounds (d) 10 pounds

Answer : b

Explanation
: Profit in first transaction = 20- 15
= 5 pounds

Profit in 2

^{nd}transaction = 40-30 = 10 pounds
Total
profit for that person = 5 + 10 = 15 pounds

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**Sample questions for Placement Papers – Solved Problems 8**

A man sold 2 pens. The initial cost of each pen was Rs. 24. If he
sells it together one at 25% profit and another 20% loss. Find the amount of
loss or gain, if he sells them separately?

Answer :

Explanation :

Answer :

Explanation :

Given CP of each pen=Rs 24

One pen is sold at 25% gain and another at 20% loss

The SP of the pen sold at profit 25% = 24 $ \times\frac{120}{100}$ = Rs 30

SP of the pen sold at 20%
loss = 24 x $ \times\frac{80}{100}$= Rs 19.2

SP of 2 pens = 30 + 19.2 =
49.2 and Total Cost price of 2 pens = 24
+ 24 =48

Total SP=30+19.2=49.2 Rs

If they sell them separately
SP =24+24=48 Rs

If they sell them separately, they will get (48 -49.6)= Rs 1.6
loss

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Raj goes to the market to buy oranges. If he can
bargain and reduce the price per orange by Rs 2 he can buy 30 oranges instead
of 20 oranges with the money he has. How much money does he have?

a)Rs 100 b) Rs 250 c) Rs 175 d) Rs 120

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**Sample questions for Placement Papers – Solved Problems 9**

The marked price of a coat was 40% less than the suggested retail price. Eesha purchased the
coat for half the marked price at the fiftieth-anniversary sale. What percentage less than the suggested retail price did Eesha pay?

a) 60 b)
20 c) 70 d) 30

Answer : C

Explanation :

Let the retail price is
Rs.100.

Then market price is 40% less than retail price.

Therefore marked price is
(100-40) % of 100 = 60.

Eesha purchased the coat
for half the marked price. = ½ x 60 =30 which is 70 less than the retail
price.

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**Sample questions for
Placement Papers – Latest Pattern Solved Problems 10**

A dealer originally bought 100 identical
batteries at a total cost of q rupees. If each battery was sold at 50 percent
above the original cost per battery, then, in terms of q, for how many rupees
was each battery sold?

a)150q b)q/150 c)3q/200 d)2q/300

Answer: C

Explanation:

Cost of 100
batteries = q rupees

Cost of 1 battery =Rs $\frac{q}{100}$

It is given that the selling price of 1 battery is 50 percent above the original cost per battery.

Cost of 1 battery =Rs $\frac{q}{100}$

It is given that the selling price of 1 battery is 50 percent above the original cost per battery.

So the selling price of 1 battery = $ \frac{q}{100}+\left[\frac{q}{100}\times\frac{50}{100}\right] $

=
$\frac{q}{100}+ \frac{q}{200} \, = \:\frac{3q}{200}$

**Sample questions for Placement Papers – New Pattern Question 11**

a)Rs 100 b) Rs 250 c) Rs 175 d) Rs 120

Answer: C

Explanation:
Let the price of one orange is x

If he can reduce the price by Rs 2 he can buy 30
oranges...

=>Total
money= 30(x -2)

Similarly, if the price is not reduced, then he
can buy 20 oranges

=>Total
money = 20x

20x =
30(x-2) => 30x -20x =60 => x =6

So total money he has => 20 x => 20 x 6 =
120